Calculate the e.f.m of the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`
[given that `E_(Cr^(3+)//Cr)^(@)=-0.75,E_(Fe^(2+)//Fe)^(@)=-0.45V]`

Calculate the e.m.f. of the cell, Cr//Cr^(3+)(0.1 M) || Fe^(2+)(0.01 M)//Fe "Given" : E_(Cr^(3+)//Cr)^(@)=-0.75" V ", E_(Fe^(2+)//Fe)^(@)=-0.45" V " "Cell reaction" : 2Cr(s)+3Fe^(2+)(aq) to 2Cr^(3+)(aq)+3Fe(s) {"Hint". E_(cell)=E_(cell)^(@)-(0.0591V)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3)}

What is the value of E^(@) cell in the following reaction? Cr|Cr^(3+) (0.1 M)||Fe^(2+) (0.01 M)|Fe Given, E_(Cr^(3+)//Cr)^(@)=-0.74 V, E_(Fe^(2+)//Fe)^(@)=-0.44 V

The EMF of the cell, Cr|Cr^(+3) (0.1M) ||Fe^(+2) (0.01M)|Fe (Given: E^(0) Cr^(+3)|Cr =- 0.75 V, E^(0) Fe^(+2)|Fe =- 0.45 V)

What is the potential for the cell Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe E^(@)Cr^(3+)// Cr=-0.74V , E^(@)Fe^(2+)//Fe=-0.44V

Find out work done for the given cell Cr| Cr^(+3)|| Fe^(+2)| Fe E_(Cr//Cr^(2+))^(@) = 0.74V, E_(Fe^(2+)//Fe)^(@) = -0.44V

Calculate the potential of the following half - cells | cells : a. Cr|Cr^(3+)(0.1 M)||Fe^(2+)(0.01M)|Fe Given : E^(c-)._(Cr^(3+)|Cr)=-0.74V E^(c-)._(Fe^(2+)|Fe)=-0.44V b. 6e^(c-)+BrO_(3)^(c-)(aq)+3H_(2)OrarrBr^(c-)(aq)+6OH(aq) Given :E^(c-)._((BrO_(3)^(c-)|Br^(c-)))=0.61V, [BrO_(3)^(c-)]=2.5xx10^(-3)M,[Br^(c-)]=5.0xx10^(-3)M,pH=9.0 c. Ag|Ag^(o+)(0.1M)||Cl^(c-)(0.02M)|Cl_(2)(g)(0.5atm)|Pt Given E_((Ag^(o+)|Ag))=0.80V,E^(c-)((Cl_(2)|2Cl^(c-))=1.36V d. NO_(3) ^(c-)(aq)+2H^(o+)(aq)+e^(-)rarrNO_(2)+H_(2)O Given : E^(c-)._(NO_(3)^(c-)|NO_(2)=0.78V What will be the reductino potential of the half cell in neutral solution ? Assuming all the other species to be at unit concentration.